Question

23-Evaluate tanA in terms of SecA
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Answer 1

Answer:

$\tan \: a = \sqrt{ {sec}^{2} \: a - 1 }$

Explanation:

$\tan \: a = \frac{ \sin \: a}{ \cos \: a}$

Squaring both sides , we get:

${ \tan }^{2} \: a = \frac{ { \sin }^{2} \: a }{ { \cos }^{2} \: a}$

We know that,

$\frac{1}{ \cos \: a} = \sec \: a$

or

$\frac{1}{ { \cos}^{2} \: a } = { \sec }^{2} \: a$

Substituting the value of cos²a in the equation, we get:

${ \tan}^{2} \: a = { \sin \:}^{2} a \times {\sec}^{2} \: a$

We also know that,

${ \sin}^{2} \: a = 1 - { \cos }^{2} \: a$

Substituting the value of sin² a in the equation, we get:

${ \tan }^{2} \: a = (1 - { \cos }^{2} \: a) \times { \sec }^{2} \: a$

Multiplying and opening bracket, we get:

${ \tan}^{2} \: a = { \sec }^{2} \: a - { \sec}^{2} \: a \times { \cos}^{2} \: a$

${ \tan}^{2} \: a = { \sec }^{2} \: a - { \sec}^{2} \: a \times \frac{1}{ { \sec}^{2} \: a}$

Multiplying and performing calculation, we get:

${ \tan }^{2} \: a = { \sec }^{2} \: a - 1$

Square rooting both sides, we get:

$\tan \: a = \sqrt{ { \sec }^{2} \: a - 1 }$

Hence Proved